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Expansion of (x+a)(x+b)

If a + b = 1,...

Question

If a + b = 1, and a, b > 0, then prove that (a + 1/a)^2 + (b + 1/b)^2 >_ 25/2

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a

b

are positive and

a + b = 1

(a + 1 / a)^{2} + (b + 1 / b)^{2} > \frac{25}{2}

x = 1 + a + a^{2} + . . . . .

y = 1 + b + b^{2} . . . .

| a | < 1

| b | < 1

1 + a b + a^{2} b^{2} + . . . . = \frac{x y}{x + y - 1}

y = \frac{2}{\sqrt{a^{2} - b^{2}}} {tan}^{- 1} (\sqrt{\frac{a - b}{a + b}} tan \frac{x}{2})

\frac{d y}{d x} = \frac{1}{a + b cos x^{'}} a > b > 0

A = (\begin{matrix} 1 & - 4 - 2 & 3 \end{matrix})

B = (\begin{matrix} - 1 & 6 3 & - 2 \end{matrix})

, then prove that

(A + B)^{2} \neq A^{2} + 2 A B + B^{2}

x = \frac{a}{a + b}

y = \frac{b}{a - b}

then prove that

\frac{1}{x} + \frac{1}{y} = \frac{a^{2} + b^{2}}{a b}

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